NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 12 Quadrilaterals Exercise 12.3

Exercise Set 12.3

Question 1. (i) If P, Q, R are the midpoints of sides AB, AC, BC respectively of △ABC, show that △PQR is congruent to △QPA and to two other triangles which you should identify.
(ii) Suppose someone erases △ABC, leaving only △PQR on the paper. Can you reconstruct △ABC from △PQR?

Answer:
(i) By the Midpoint Theorem:
1. P, Q midpoints of AB, AC: PQ = \( \frac{1}{2} \)BC = BR = RC.
2. Q, R midpoints of AC, BC: QR = \( \frac{1}{2} \)AB = AP = PB.
3. P, R midpoints of AB, BC: PR = \( \frac{1}{2} \)AC = AQ = QC.
△PQR and △QPA (P ↔ Q, Q ↔ P, R ↔ A): PQ = QP, QR = PA, RP = AQ. So △PQR ≅ △QPA (SSS).
△PQR and △RBP (P ↔ R, Q ↔ B, R ↔ P): PQ = RB, QR = BP, RP = PR. So △PQR ≅ △RBP (SSS).
△PQR and △CRQ (P ↔ C, Q ↔ R, R ↔ Q): PQ = CR, QR = RQ, RP = QC. So △PQR ≅ △CRQ (SSS).
Hence △PQR is congruent to △QPA, △RBP and △CRQ: all four small triangles are congruent.
(ii) Yes. By the Midpoint Theorem each side of △ABC passes through one vertex of △PQR and is parallel to the opposite side of △PQR:
side AB passes through P and is parallel to QR; side AC passes through Q and is parallel to PR; side BC passes through R and is parallel to PQ.
Construction: through P draw the line parallel to QR, through Q the line parallel to PR, and through R the line parallel to PQ. These lines meet in pairs at A, B and C.

This really is the erased triangle: there is only one line through P parallel to QR, and line AB is such a line, so the drawn line is AB. The same holds for AC and BC, so their meeting points are A, B, C.

Teacher's Note:
1. Match vertices carefully before writing each congruence.
2. Part (ii) uses the uniqueness of the parallel through a point.

Question 2. In △ABC, let M and N be midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD.

Answer:
Let MN meet AD at X. We need AX = XD.
1. M and N are midpoints of AB and AC, so by the Midpoint Theorem MN ∥ BC. Therefore MX ∥ BD.
2. In △ABD, M is the midpoint of AB and the line MX is parallel to BD.
3. By Theorem 7 (converse of the Midpoint Theorem), this line bisects the third side AD. So X is the midpoint of AD: AX = XD.
Hence MN bisects AD, wherever D lies on BC.

Question 3. In a quadrilateral ABCD, suppose AB ∥ DC and AB ≠ CD. Suppose G and H are the midpoints of AC and BD respectively. Prove that GH ∥ AB. (Why did we assume AB ≠ CD?)

Answer:
Construction: let M be the midpoint of AD.
1. In △ADC, M and G are midpoints of AD and AC, so MG ∥ DC (Midpoint Theorem).
2. In △DAB, M and H are midpoints of DA and DB, so MH ∥ AB.
3. As AB ∥ DC, both MG and MH are parallel to AB and pass through M. Only one line through M is parallel to AB, so M, G, H lie on it.
Hence line GH is this line, and GH ∥ AB (∥ DC).
Why AB ≠ CD? If AB = CD as well as AB ∥ DC, then ABCD is a parallelogram (Theorem 5). Its diagonals bisect each other, so G and H are the same point. There is then no segment GH and 'GH ∥ AB' makes no sense.
(Also MG = \( \frac{1}{2} \)DC and MH = \( \frac{1}{2} \)AB, so GH = \( \frac{1}{2} \)|DC − AB|, which is zero exactly when AB = CD.)

Teacher's Note:
1. The midpoint M of AD links the two triangles.
2. Uniqueness of the parallel through M makes G, H and M collinear.

Question 4. Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively.
(i) Show that PR and QS bisect each other.
(ii) Show that if AC = BD, then PR and QS are perpendicular. Is the converse true?

Answer:
(i) By Theorem 9, PQRS is a parallelogram (the Varignon parallelogram). PR and QS are its diagonals, and the diagonals of a parallelogram bisect each other (Theorem 1(c)). So PR and QS bisect each other (at O).
(ii) By the Midpoint Theorem, PQ = \( \frac{1}{2} \)AC (in △ABC) and QR = \( \frac{1}{2} \)BD (in △BCD).
If AC = BD, then PQ = QR. A parallelogram with two adjacent sides equal is a rhombus, and the diagonals of a rhombus are perpendicular. So PR ⊥ QS.
Converse: yes, it is true. If PR ⊥ QS, then PQRS is a parallelogram whose diagonals bisect each other at right angles, so it is a rhombus (Exercise Set 12.2, Q3(ii)).
Then PQ = QR, i.e. \( \frac{1}{2} \)AC = \( \frac{1}{2} \)BD, so AC = BD.
So PR ⊥ QS if and only if AC = BD.

Question 5. Suppose PQRS is the Varignon parallelogram of ABCD.
(i) Copy only PQRS on another paper. Show how you will recreate a congruent copy A′B′C′D′ of ABCD from PQRS. This will be relevant when we return to study tilings at the end of this chapter. (Hint: How will you place vertex A′? How will you place B′, C′ and D′?)
(ii) There are multiple ways to construct A′B′C′D′ in (i). Justify why the quadrilateral A′B′C′D′ you constructed is congruent to ABCD. You may need to show why S is collinear with the points A′ and D′ that you constructed, and similarly for P, Q and R. (Hint: Use congruence of triangles, for example △SDR ≅ △SD′R.)
(iii) Show that if PQRS is a square, then AC and BD are perpendicular and equal. Prove the converse.

Answer:
(i) Construction (measurements are taken from the original ABCD):
1. Place A′. On the side of PS away from Q and R, build △A′PS congruent to the corner triangle APS: arc of radius AP about P, arc of radius AS about S; they cross at A′.
2. Place B′. Extend A′P beyond P to B′ with PB′ = PA′ (so P is the midpoint of A′B′).
3. Place C′. Extend B′Q beyond Q to C′ with QC′ = QB′.
4. Place D′. Extend C′R beyond R to D′ with RD′ = RC′.
5. Join D′A′. It passes through S, and A′B′C′D′ is the required copy.

(Another way: copy each of the four corner triangles A′PS, B′QP, C′RQ, D′SR of ABCD separately onto the four sides of PQRS.)
(ii) S is the midpoint of D′A′. By construction P, Q, R are the midpoints of A′B′, B′C′, C′D′. Let S′ be the midpoint of D′A′. By Theorem 9, PQRS′ is a parallelogram. Only one point completes P, Q, R to a parallelogram PQRS′ (through P draw the parallel to QR, through R the parallel to QP). PQRS is such a parallelogram, so S′ = S. Hence A′, S, D′ are collinear.
Corner triangles are congruent. By construction △A′PS ≅ △APS, so A′S = AS and ∠A′SP = ∠ASP.
Along the straight lines A′SD′ and ASD: ∠A′SP + ∠PSR + ∠RSD′ = 180° and ∠ASP + ∠PSR + ∠RSD = 180°. So ∠RSD′ = ∠RSD.
Also SD′ = SA′ = SA = SD. In △SDR and △SD′R: SD = SD′, ∠RSD = ∠RSD′, SR common, so △SDR ≅ △SD′R (SAS). This gives RD′ = RD and ∠SRD′ = ∠SRD.
Repeating the argument at R (RC′ = RD′ = RD = RC) gives △RC′Q ≅ △RCQ, and at Q gives △QB′P ≅ △QBP.
So A′B′C′D′ is the same parallelogram PQRS with four corner triangles congruent to those of ABCD, attached in the same way. All corresponding sides and angles match (for example A′B′ = 2PA′ = 2PA = AB), so A′B′C′D′ ≅ ABCD.
(iii) By the Midpoint Theorem, PQ ∥ AC, PQ = \( \frac{1}{2} \)AC and QR ∥ BD, QR = \( \frac{1}{2} \)BD.
If PQRS is a square, PQ ⊥ QR. Since AC ∥ PQ and BD ∥ QR, we get AC ⊥ BD. Also PQ = QR, so AC = 2PQ = 2QR = BD: AC = BD.
Converse: let AC ⊥ BD and AC = BD. PQRS is a parallelogram (Theorem 9). AC = BD gives PQ = QR, so it is a rhombus; AC ⊥ BD gives PQ ⊥ QR, so ∠PQR = 90°. A rhombus with a right angle is a square.

Teacher's Note:
1. Treat ABCD as 'PQRS plus four corner triangles'; the proof copies them one by one.
2. In (iii), lengths of PQRS come from the diagonal lengths and its angles from the angle between the diagonals.